PROBLEMS ON AVERAGE
Problems on Average :
In this section, you will learn how average problems can be solved easily.
The formula given below can be used to find average of the given values.
Problems on Average - Examples
Example 1 :
Find the average of first 20 natural numbers which are divisible by 7.
Solution :
Step 1 :
The first natural number which is divisible by 7 is 7.
The next numbers which are divisible by 7 are 14, 21.....
Step 2 :
Write the first twenty natural numbers which are divisible by 7.
They are 7, 14, 21, 28........ up to 20 terms.
Step 3 :
Find the sum of all the above numbers.
= 7 + 14 + 21 + 28.........up to 20 term
Because all of the above numbers are divisible by 7, we can factor 7.
= 7(1 + 2 + 3 + 4 +.........+ 20)
= 7 ⋅ 210
= 1470
Step 4 :
Find the average.
Average = Sum of all 20 numbers / 20
Average = 1470 / 20
Average = 73.5
Hence average of first 20 natural numbers which are divisible by 7 is 73.5
Example 2 :
The average of four consecutive even numbers is 27. Find the largest of these numbers.
Solution :
Step 1 :
Let "x' be the first even number.
Then the four consecutive even numbers are
x, x + 2, x + 4, x + 6
Step 2 :
Average of the four consecutive even numbers is 27.
So, we have
(x + x + 2 + x + 4 + x + 6) / 4 = 27
(4x + 12) = 108
4x = 96
x = 24
Then, the value of the largest number is
x + 6 = 24 + 6 = 30
Hence, the largest of the four consecutive even numbers is 30.
Example 3 :
There are two sections A and B of a class, consisting 36 and 44 students respectively. If the average weight of the section A is 40 kg and that of section B is 35 kg, find the average weight of the whole class.
Solution :
Step 1 :
For section A, average weight is 40 kg.
That is,
Sum of the weights of 36 students / 36 = 40
Multiply both sides by 36.
Sum of the weights of 36 students = 40 ⋅ 36
Sum of the weights of 36 students = 1440
Step 2 :
For section B, average weight is 35 kg.
That is,
Sum of the weights of 44 students / 44 = 35
Multiply both sides by 44.
Sum of the weights of 44 students = 35 ⋅ 44
Sum of the weights of 44 students = 1540
Step 3 :
Total weight of 80 students (whole class) is
= 1440 + 1540
= 2980
Step 4 :
Find the average weight of the whole class.
Average weight of the whole class is
= Total weight / No. of students
= 2980 / 80
= 37.25 kg
Hence, the average weight of the whole class is 37.25 kg.
Example 4 :
In John's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doesn't agree with John and he thinks that John's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of John ?
Solution :
Let "x" be John's weight
Step 1 :
According to John, we have 65 < x < 72.
According to his brother, we have 60 < x < 70.
According to his mother, we have x ≤ 68.
Step 2 :
The values of "x" which satisfy all the above three conditions are 66, 67 and 68.
Step 3 :
Average of the above three values is
= Sum of the three values / 3
= (66 + 67 + 68) / 3
= 201/3
= 67
Hence, the average of different probable weights of John is 67 kg.
Example 5 :
A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find the average score average after 16th match.
Solution :
Step 1 :
Let "x" be the average after 16th match.
Then, the average after 17th match is (x + 3).
Step 2 :
Average after 17 matches = x + 3
Total runs scored in 17 matches / 17 = x + 3
Total runs scored in 17 matches = 17(x + 3)
Total runs scored in 17 matches = 17x + 51 ----- (1)
Step 3 :
Average after 16 matches = x
Total runs scored in 16 matches / 16 = x
Total runs scored in 16 matches = 16x
Given : Runs cored in 17th match = 87
Then, total runs scored in 17 matches = 16x + 87 -----(2)
Step 4 :
From (1) and (2),
17x + 51 = 16x + 87
x + 51 = 87
Subtract 51 from both sides.
x = 36
Hence, the average score after 16th match is 36.
After having gone through the stuff given above, we hope that the students would have understood how to solve problem on average.
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